package com.shuang.dp38;
// dp数组中存储word1和word2最长相同子序列的长度
// class Solution {
//     public int minDistance(String word1, String word2) {
//         int len1 = word1.length();
//         int len2 = word2.length();
//         int[][] dp = new int[len1 + 1][len2 + 1];

//         for (int i = 1; i <= len1; i++) {
//             for (int j = 1; j <= len2; j++) {
//                 if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
//                     dp[i][j] = dp[i - 1][j - 1] + 1;
//                 } else {
//                     dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
//                 }
//             }
//         }

//         return len1 + len2 - dp[len1][len2] * 2;
//     }
// }

// dp 以i-1为结尾和j-1为结尾的  需要删除操作的次数
class Solution {
    public int minDistance(String word1, String word2) {
        int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        for (int i = 0; i < word1.length() + 1; i++)
            dp[i][0] = i;
        for (int j = 0; j < word2.length() + 1; j++)
            dp[0][j] = j;

        for (int i = 1; i < word1.length() + 1; i++) {
            for (int j = 1; j < word2.length() + 1; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = Math.min(dp[i - 1][j - 1] + 2,
                            Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
                }
            }
        }

        return dp[word1.length()][word2.length()];
    }
}